Question: Solve for $t$, $ -\dfrac{4t - 6}{3t^3} = \dfrac{1}{12t^3} + \dfrac{5}{3t^3} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3t^3$ $12t^3$ and $3t^3$ The common denominator is $12t^3$ To get $12t^3$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{4t - 6}{3t^3} \times \dfrac{4}{4} = -\dfrac{16t - 24}{12t^3} $ The denominator of the second term is already $12t^3$ , so we don't need to change it. To get $12t^3$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{5}{3t^3} \times \dfrac{4}{4} = \dfrac{20}{12t^3} $ This give us: $ -\dfrac{16t - 24}{12t^3} = \dfrac{1}{12t^3} + \dfrac{20}{12t^3} $ If we multiply both sides of the equation by $12t^3$ , we get: $ -16t + 24 = 1 + 20$ $ -16t + 24 = 21$ $ -16t = -3 $ $ t = \dfrac{3}{16}$